If ABCD is a rectangle, what is the area of the shaded region?

Question

https://gmatclub.com/forum/download/file.php?id=25347 If ABCD is a rectangle, what is the area of the shaded region? A. 64 B. 82 C. 96 D. 120 E. 150 Untitled.png

PareshGmat · Accepted Answer

Answer = C. 96 Untitled.png Diagonal of rectangle = \sqrt{20^2 + 15^2} = \sqrt{625} = 25 Half the area of rectangle = 150 = \frac{1}{2} * 25 * h Height = 12 = CP Length BP = \sqrt{20^2 - 12^2} = \sqrt{256} = 16 Area of shaded region riangle BPC = \frac{1}{2} * 16 * 12 = 96

EMPOWERgmatRichC · Answer

Hi All,

When the GMAT puts a right triangle in front of you, it is probably NOT a random right triangle. While there are some exceptions, in most cases, you're looking at a 3/4/5, 5/12/13, 30/60/90 or 45/45/90 right triangle. If you see ANY clues that point to a specific right triangle, then you should think about how you can prove that the triangle IS what you think it is.

Here, we have 3 right triangles and they all appear to be multiples of a 3/4/5 (notice how each has a diagonal that is a MULTIPLE of 5).

Assuming they're all 3/4/5s, the sides SHOULD be....

Big triangle: 15/20/25
Middle triangle: 12/16/20
Small triangle: 9/12/15

From the picture, there are a couple of other things that need to "line up" to prove that we're correct:

1) One side of the "small" and "middle" triangles MUST be the same.
2) The "other" sides of each of those triangles MUST add up to 25 (the diagonal of the big triangle)

The common "shared" side would be 12, which means the "other" sides (16 and 9) add up to 25. So we have the EXACT situation that we thought we did. With knowledge of ALL of the side lengths, we can now answer any question that's asked.

GMAT assassins aren't born, they're made,
Rich

sunita123 · Answer

Area of Triangle BCD =1/2*20*15=150

BD= sqrt (20^2+15^2)=25
also area of BCD can be determined as 1/2* BD* height from C to BD
lets height from C meets BD at X 
 Area of triangle BCX will be
1/2* BD* CX =150
=>1/2*25* CX=150
=>CX=300/25=12

Now, BX= sqrt (20^2-12^2)=16
So, area of shaded region will be
1/2* BX*CX=1/2*16*12=96

Answer is C

chetan2u · Answer

a way different from the one mentioned ....ans C
 we can use similar triangles property between triangles abd and bcp and get the sides of triangle bcp and so area comes as 96..

DesiGmat · Answer

Here we go:

Consider triangle BCD

Area will be
Let the perpendiculer from C to BD be CM

1/2 * (CD*BC) = 1/2 * (CM * BD)

Solve

150 = 1/2 * (CM * 25) (BD is the diagonal---> Apply pythogorus Theorem)

CM = 12

Again apply pythagorus in shaded triangle

BM = 16

So Area will be 1/2 * (BM * CM) = 96

Option C is correct

vipulgoel · Answer

If ABCD is a rectangle, what is the area of the shaded region?

A. 64
B. 82
C. 96
D. 120
E. 150

Kudos for a correct solution.

Pythagorean Theorem to solve for BD = 20^2 + 15^2 = 225 + 400 = 625
BD = sqrt (625) = 25

All these triangles inscribed are proportionate, since BD is a diagonal of this rectangle
Since BC = 20 and BD = 25, BC = = 4/5*BD
So BX = 4/5 * 20 = 16
Solve for CX = A^2 + B^2 = C^2
256 + B^2 = 400
B^2 = 400-256
B^2 = 144
Sqrt(144) = 12

Area of shaded region = 12*16 / 2 = 12 * 8 = 96

Hence answer C

bluesquare · Answer

The problem with your solution is that angle CXD and XCB are not 45 degrees - if they were, CX would bisect BD, but it does not (unless ABCD were a rhombus).

The correct solution: We are looking for  \frac{CX*BX}{2} 
We know: BD is 25 (3-4-5 right triangle * 5), Area of BCD = 150, and therefore CX = 12 (because  \frac{CX*BD}{2} = 150 )
So therefore:  BX^2 + 144 = 400 ->  BX^2 = 256  ->  BX = 16  -> Area is  \frac{16*12}{2} = 96